Respuesta :
Answer:
a) 0.1829
b) 0.6823
c) 0.0413
Step-by-step explanation:
We are given the following information:
We treat adult having little confidence in the newspaper as a success.
P(Adult have little confidence) = 62% = 0.62
Then the number of adults follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 10
a) exactly 5
[tex]P(x =5)\\\\=\binom{10}{5}(0.62)^5(1-0.62)^5\\\\=0.1829[/tex]
0.1829 is the probability that exactly 5 out of 10 U.S.adults have very little confidence in newspapers.
b) atleast six
[tex]P(x \geq 6)\\\\ = P(x = 6) + P(x = 7)+P(x = 8)+P(x = 9)+P(x=10)\\\\= \displaystyle\sum \binom{10}{n}(0.62)^n(1-0.62)^{10-n}, n =6,7,8,9,10\\\\= 0.6823[/tex]
0.6823 is the probability that atleast 6 out of 10 U.S. adults have very little confidence in newspapers.
c) less than four
[tex]P(x <4)\\\\ = P(x = 0) + P(x = 1)+P(x = 2)+P(x = 3)\\\\= \displaystyle\sum \binom{10}{n}(0.62)^n(1-0.62)^{10-n}, n =0,1,2,3\\\\= 0.0413[/tex]
0.0413 is the probability that less than 4 out of 10 U.S. adults have very little confidence in newspapers.
